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E cross drift calculus

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data

    good afternoon,

    I am a problem with the E cross B drift velocity calculus from a charged particle. At first, an eletron is submitted a uniform magnetic field and have a circular trajetory. After, appear a uniform eletric field that create a drift velocity. The equations are:

    m dv/dt = qE + qv X B.

    Taking v constant or the mean aceleration zero, arrive the equation:

    v = (E X B)/B^2

    But I don't understand why the v is constant or mean aceleration is zero. Anyone can help me?
  2. jcsd
  3. Sep 28, 2015 #2
    I think you first have to realise that you are dealing with vector quantities.

    With just a magnetic field of strength B you have the force acting on a charge q moving with velocity v is mdv/dt = q vXB (where vectors are bold face).

    In this situation if you assume that the charge moves perpendicular to the magnetic field (what does that mean for the term vXB?) then, as you have stated, the charge moves in a circular path.

    Question: What if the charge of the particle has a component of its velocity in the same direction as the magnetic field?

    When you apply an electric field, E, the total force acting on the charge is mdv/dt = qE + q vXB.

    The velocity you determine (making use of vector notation) v = EXB/(B.B) is the condition for dv/dt = 0 and its not clear to me whether you have presented all the information?
  4. Sep 28, 2015 #3
    sorry for I don't put in vector notation, but I write thinking in vectors. The problem is why dv/dt=0?
  5. Sep 28, 2015 #4
    No problem about the vector notation.

    The reason I wondered if there was more to the question because, in general, dv/dt wouldn't be zero.

    Look up "velocity selector", for example https://en.wikipedia.org/wiki/Wien_filter (must admit I learned something as well, I never realised that velocity selectors in this case were referred to as Wien filters).

    If you look at the attached figure (taken from Wikipedia) you can see that the choice v = EXB/(B.B) is quite special.

    velocity selector.png
  6. Sep 28, 2015 #5
    I think nobody erason for dv/dt=0 too. But, in all places that I read they use this condition. Is it only a hipothesis? I want to know if it was a hypothesis or something general of the problem.
  7. Sep 28, 2015 #6
    Something else you need to consider: To derive the expression for v you used the result
    vXB = - E
    then I guess you took the cross product with B giving
    BX(vXB) = -BXE = EXB
    Using the rules for vector cross product gives the left hand side as
    v B2 + B v.B
    so that you can only determine v for the case v.B = 0, namely that v and B are perpendicular
  8. Sep 28, 2015 #7
    v is the perpendicular component of velocity related tho magnetic field.
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